SOLUTION: hey i would really appreciate it if you could solve this combinatorics problem for me: P(n,3)= 2C(n,2), nϵN.

Algebra ->  Test -> SOLUTION: hey i would really appreciate it if you could solve this combinatorics problem for me: P(n,3)= 2C(n,2), nϵN.      Log On


   

Question 154622: hey i would really appreciate it if you could solve this combinatorics problem for me:
P(n,3)= 2C(n,2), nϵN.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
P(n,3)= 2C(n,2)
----------------------
n!/(n-3)! = 2[n!/(n-2)!2!]
Cancel the n! and the 2, both of which are common factors to get:
1/(n-3)! = 1/(n-2)!
Invert to get:
(n-3)! = (n-2)!
(n-3)! = (n-2)(n-3)!
Divide both sides by (n-3)!
n-2 = 1
n = 3
=========================
Cheers,
Stan H.